Javascript 刷新JustGage而不重新加载页面
当我在同一页面上发布表单输入的数字时,我试图在不重新加载整个页面的情况下进行刷新Javascript 刷新JustGage而不重新加载页面,javascript,php,jquery,justgage,Javascript,Php,Jquery,Justgage,当我在同一页面上发布表单输入的数字时,我试图在不重新加载整个页面的情况下进行刷新 <div id="gauge" class="400x260px"></div> <!---The Form---> <form class="form-inline" action="" id="myform" form="" method="post"> <!-- Text input--> <div class="form-group"&g
<div id="gauge" class="400x260px"></div>
<!---The Form--->
<form class="form-inline" action="" id="myform" form="" method="post">
<!-- Text input-->
<div class="form-group">
<label class="col-md-4 control-label" for="total"></label>
<div class="col-md-8">
<input id="total" name="total" type="text" placeholder="total" class="form-control input-lg" required>
</div>
</div>
<!-- Button -->
<div class="form-group">
<label class="col-md-4 control-label" for="submit1"></label>
<div class="col-md-4">
<button id="submitButtonId" name="submit1" class="btn btn-primary btn-xl">Submit</button>
</div>
</div>
</form>
提交
jQuery:
$(document).ready(function(){
//JSON REQUEST
$.getJSON("test_server.php",function(result){
var g = new JustGage({
id: "gauge",
value: result.total,
min: 0,
max: 100,
title: "Data from Form"
});
});
//Post Form
$("#submitButtonId").on("click",function(e){
e.preventDefault();
var formdata = $(this.form).serialize();
$.post('test_server.php', formdata, function(data){
//Reset Form
$('#myform')[0].reset();
});
return false;
});
});
</script>
$(文档).ready(函数(){
//JSON请求
$.getJSON(“test_server.php”,函数(结果){
var g=新的JustGage({
id:“仪表”,
值:result.total,
分:0,,
最高:100,
标题:“表格中的数据”
});
});
//邮政表格
$(“#提交按钮ID”)。在(“单击”上,函数(e){
e、 预防默认值();
var formdata=$(this.form).serialize();
$.post('test_server.php',formdata,function(data){
//重置表单
$('#myform')[0].reset();
});
返回false;
});
});
test_server.php:
<?php
$total= mysqli_real_escape_string($db_conx,$_POST['total']);
$my_data=array(total=>"$total");
// sending output
header('Content-Type: text/json');
echo json_encode($my_data,true);
?>